Hello, Punch!
I assume there are more parts to this problem.
There is way too much information . . .
In the diagram, $\displaystyle \Delta ACD$ has points: $\displaystyle A(2,12)$ and $\displaystyle B(6,9).$
$\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC $.
The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$.
The equation of $\displaystyle AD$ is: $\displaystyle 3y\:=\:5x+26$.
Calculate the coordinates of $\displaystyle C.$
Here is a simplified diagram:
Code:

 A
 (2,12)o   +
 \ :
 \ :
 \ :
 \ :
 \:
 B(6,9)o     +
 \ :
 \ :
 \ :
 \ :
 \ :
 \ :
 \ :
 \ :
 \:
 C o

  +              

$\displaystyle \text{Since }BC = \frac{2}{3}AC,\; BC\text{ is twice }AB$
Going from $\displaystyle A$ to $\displaystyle B$, we move: 4 units right and 3 units down.
So, going from $\displaystyle B$ to $\displaystyle C$, we move: 8 units right and 6 units down.
Therefore, point $\displaystyle C$ is $\displaystyle (14,3).$